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Musya8 [376]
3 years ago
10

Suppose A is n x n matrix and the equation Ax = 0 has only the trivial solution. Explain why A has n pivot columns and A is row

equivalent to In. By Theorem 7, this shows that A must be Invertible.)
Theorem 7: An n x n matrix A is invertible if and only if A is row equivalent to In, and in this case, any sequence of elementary row operations that reduces A to In also transfrms In into A-1.
Mathematics
1 answer:
vovangra [49]3 years ago
4 0

Answer:

Remember, a homogeneous system always is consistent. Then we can reason with the rank of the matrix.

If the system Ax=0 has only the trivial solution that's mean that the echelon form of A hasn't free variables, therefore each column of the matrix has a pivot.

Since each column has a pivot then we can form the reduced echelon form of the A, and leave each pivot as 1 and the others components of the column will be zero. This means that the reduced echelon form of A is the identity matrix and so on A is row equivalent to identity matrix.

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The functions are inverses; f(g(x)) = x ⇒ answer D

h^{-1}(x)=\sqrt{\frac{x+1}{3}} ⇒ answer D

Step-by-step explanation:

* <em>Lets explain how to find the inverse of a function</em>

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = f^{-1}(x)

* <em>Lets use these steps to solve the problems</em>

∵ f(x)=\sqrt{x-3}

∵ f(x) = y

∴ y=\sqrt{x-3}

- Exchange x and y

∴ x=\sqrt{y-3}

- Square the two sides

∴ x² = y - 3

- Add 3 to both sides

∴ x² + 3 = y

- Change y by f^{-1}(x)

∴ f^{-1}(x)=x^{2}+3

∵ g(x) = x² + 3

∴ f^{-1}(x)=g(x)

∴ <u><em>The functions are inverses to each other</em></u>

* <em>Now lets find f(g(x))</em>

- To find f(g(x)) substitute x in f(x) by g(x)

∵ f(x)=\sqrt{x-3}

∵ g(x) = x² + 3

∴ f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x

∴ <u><em>f(g(x)) = x</em></u>

∴ The functions are inverses; f(g(x)) = x

* <em>Lets find the inverse of h(x)</em>

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

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- Divide both sides by 3

∴ \frac{x + 1}{3}=y^{2}

- Take √ for both sides

∴ ± \sqrt{\frac{x+1}{3}}=y

∵ x ≥ 0

∴ We will chose the positive value of the square root

∴ \sqrt{\frac{x+1}{3}}=y

- replace y by h^{-1}(x)

∴ h^{-1}(x)=\sqrt{\frac{x+1}{3}}

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